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In this question, the OP asks to find the number of divisors of $2^2\cdot 3^3\cdot 5^3\cdot 7^5$ which are of the form $4n+1,n\in N$. The top answer points out that the required divisor is of the form $$3^a\cdot 5^b\cdot 7^c$$ with $0\leq a\leq 3,0\leq b\leq 3,0\leq c\leq 5$ and $a+c$ being even. The answer therefore is, apparently, $(4 \cdot 4 \cdot 6)/2=48$.

But this is wrong according to my book: the correct answer is $47$. Obviously, one case has been overcounted, but which? As far as I know, the person who wrote the top answer employed a fairly standard approach and should have arrived at the correct answer.

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    $\begingroup$ I guess the book uses $0 \notin \mathbb{N}$, so it doesn't count $1 = 4\cdot 0 + 1$. $\endgroup$ – Daniel Fischer♦ Nov 20 '20 at 16:18
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    $\begingroup$ Well, at a guess, the official solution excludes $(a,b,c)=(0,0,0)$, i.e. it excludes the divisor $1$. $\endgroup$ – lulu Nov 20 '20 at 16:18
  • $\begingroup$ @Daniel Fischer Ah, that makes sense. Thanks. $\endgroup$ – Ray Bradbury Nov 20 '20 at 16:21
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    $\begingroup$ Does this answer your question? Number of divisors of the form $(4n+1)$ $\endgroup$ – Dave Nov 22 '20 at 7:41
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As pointed out in the comments by Daniel Fischer and lulu, my book considers $0 \notin N$, so discounts the case where $a=b=c=0$, i.e., $4(0)+1=1$.

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